course Phy 121 My assignment 30 has been posted on my access page but there is not a note at the end of it. It was posted on April 13.
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08:19:55 `q001. Note that this assignment contains 3 questions. The moment of inertia of a concentrated mass m lying at a distance r from the axis of rotation is m r^2. Moments of inertia are additive--that is, if an object with a moment of inertia about some axis is added to another object with its moment of inertia about the same axis, the moment of inertia of the system about that axis is found by simply adding the moments of inertia of the two objects. Suppose that a uniform steel disk has moment of inertia .0713 kg m^2 about an axis through its center and perpendicular to its plane. If a magnet with mass 50 grams is attached to the disk at a point 30 cm from the axis, what will be the moment of inertia of the new system?
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RESPONSE --> Mass had moment of inertia I = m r^2 = .05 kg * (.30 m)^2 = .0045 kg m^2. The moment of inertia of the new system = .0713 kg m^2 + .0045 kg m^2 = .0758 kg m^2
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08:20:02 A mass of m = .05 kg at distance r = .30 meters from the axis of rotation has moment of inertia I = m r^2 = .05 kg * (.30 m)^2 = .0045 kg m^2. The moment of inertia of the new system will therefore be the sum .0713 kg m^2 + .0045 kg m^2 = .0758 kg m^2 of the moments of inertia of its components, the disk and the magnet.
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RESPONSE --> ok
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08:27:23 `q002. A uniform rod with mass 5 kg is 3 meters long. Masses of .5 kg are added at the ends and at .5 meter intervals along the rod. What is the moment of inertia of the resulting system about the center of the rod?
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RESPONSE --> The rod moment of inertia = 1/12 M L^2 = 1/12 * 5 kg * (3 m)^2 = 3.75 kg m^2 1.5 m from the center moment of inertia = m r^2 = .5 kg * (1.5 m)^2 = 1.125 kg m^2 total moment of inertia is 2.25 kg m^2 1 m from the center moment inertia = m r^2 = .5 kg * (1 m)^2 = .5 kg m^2 total moment of intertia = 1 kg m^2 .5 m from the center moment of inertia = m r^2 = .5 kg ( .5 m)^2 = .125 kg m^2 total = .25 kg m^2 The total moment of inertia of the added masses is 2.25 kg m^2 + 1 kg m^2 + .25 kg m^2 = 3.5 kg m^2 Add this to the moment of inertia of the rod and total moment of inertia is 3.75 kg m^2 + 3.5 kg m^2 = 7.25 kg m^2
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08:29:23 The rod itself, being rotated about its center, has moment of inertia 1/12 M L^2 = 1/12 * 5 kg * (3 m)^2 = 3.75 kg m^2. The added masses are at distances 1.5 meters (the two masses masses on the ends), 1.0 meters (the two masses .5 m from the ends), .5 meters (the two masses 1 m from the ends) and 0 meters (the mass at the middle of the rod) from the center of the rod, which is the axis of rotation. At 1.5 m from the center a .5 kg mass will have moment of inertia m r^2 = .5 kg * (1.5 m)^2 = 1.125 kg m^2; there are two such masses and their total moment of inertia is 2.25 kg m^2. The two masses lying at 1 m from the center each have moment inertia m r^2 = .5 kg * (1 m)^2 = .5 kg m^2, so the total of the two masses is double is, or 1 kg m^2. {}The two masses lying at .5 m from the center each have moment of inertia m r^2 = .5 kg ( .5 m)^2 = .125 kg m^2, so their total is double this, or .25 kg m^2. The mass lying at the center has r = 0 so m r^2 = 0; it therefore makes no contribution to the moment of inertia. The total moment of inertia of the added masses is therefore 2.25 kg m^2 + 1 kg m^2 + .25 kg m^2 = 3.5 kg m^2. Adding this to the he moment of inertia of the rod itself, total moment of inertia is 3.75 kg m^2 + 3.5 kg m^2 = 7.25 kg m^2. We note that the added masses, even including the one at the center which doesn't contribute to the moment of inertia, total only 3.5 kg, which is less than the mass of the rod; however these masses contribute as much to the moment of inertia of the system as the more massive uniform rod.
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RESPONSE --> ok
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08:34:11 `q003. A uniform disk of mass 8 kg and radius .4 meters rotates about an axis through its center and perpendicular to its plane. A uniform rod with mass 10 kg, whose length is equal to the diameter of the disk, is attached to the disk with its center coinciding with the center of the disk. The system is subjected to a torque of .8 m N. What will be its acceleration and how to long will it take the system to complete its first rotation, assuming it starts from rest?
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RESPONSE --> The moment of inertia of the disk is 1/2 M R^2 = 1/2 * 8 kg * (.4 m)^2 = .64 kg m^2 The rod's moment of inertia will be 1/12 M L^2 = 1/12 * 10 kg * (.8 m)^2 = .53 kg m^2 The total moment of inertia of the system is .64 kg m^2 + .53 kg m^2 = 1.17 kg m^2 a = `alpha = `tau / I = .8 m N / (1.17 kg m^2) = .7 rad/s^2 `ds = v0 `dt + 1/2 a `dt^2, which for v0=0 is `ds = 1/2 a `dt^2, write in terms of the angular quantities `d`theta = 1/2 `alpha `dt^2 `dt = +- `sqrt( 2 `d`theta / `alpha ) = +- `sqrt( 2 * 2 `pi rad / (.7 rad/s^2)) = +-`sqrt( 12.56 rad / (.7 rad/s^2) ) = +-4.2 sec
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08:35:02 The moment of inertia of the disk is 1/2 M R^2 = 1/2 * 8 kg * (.4 m)^2 = .64 kg m^2. The rod will be rotating about its center so its moment of inertia will be 1/12 M L^2 = 1/12 * 10 kg * (.8 m)^2 = .53 kg m^2 (approx). ( Note that the rod, despite its greater mass and length equal to the diameter of the disk, has less moment of inertia. This can happen because the mass of the disk is concentrated more near the rim than near the center (there is more mass in the outermost cm of the disk than in the innermost cm), while the mass of the rod is concentrated the same from cm to cm. ). The total moment of inertia of the system is thus .64 kg m^2 + .53 kg m^2 = 1.17 kg m^2. The acceleration of the system when subject to a .8 m N torque will therefore be `alpha = `tau / I = .8 m N / (1.17 kg m^2) = .7 rad/s^2, approx.. To find the time required to complete one revolution from rest we note that the initial angular velocity is 0, the angular displacement is 1 revolution or 2 `pi radians, and the angular acceleration is .7 rad/s^2. By analogy with `ds = v0 `dt + 1/2 a `dt^2, which for v0=0 is `ds = 1/2 a `dt^2, we write in terms of the angular quantities `d`theta = 1/2 `alpha `dt^2 so that `dt = +- `sqrt( 2 `d`theta / `alpha ) = +- `sqrt( 2 * 2 `pi rad / (.7 rad/s^2)) = +-`sqrt( 12.56 rad / (.7 rad/s^2) ) = +-4.2 sec. We choose the positive value of `dt, obtaining `dt = +4.2 sec..
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RESPONSE --> ok
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assignment #032 [뺗܃ܝ Physics I 04-14-2006
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08:38:04 Query experiment to be viewed. What part or parts of the system experiences a potential energy decrease? What part or parts of the system experience(s) a kinetic energy increase?
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RESPONSE --> The mass on the string descends and loses PE. The wheel and the descending mass increase in KE, along with the other smaller and slower-moving parts.
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08:38:10 ** The mass on the string descends and loses PE. The wheel and the descending mass both increase in KE, as do the other less massive parts of the system (e.g., the string) and slower-moving parts (e.g., the axel, which rotates at the same rate as the wheel but which due to its much smaller radius does not move nearly as fast as most of the wheel). **
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RESPONSE --> ok
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08:38:40 What part or parts of the system experience(s) an increase in angular kinetic energy?
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RESPONSE --> The wheel, the bolts, the axle, and anything else that's rotating experiences an increase in angular KE.
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08:38:46 ** The wheel, the bolts, the axle, and anything else that's rotating experiences an increase in angular KE. **
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RESPONSE --> ok
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08:39:09 What part or parts of the system experience(s) an increasing translational kinetic energy?
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RESPONSE --> The descending mass.
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08:39:13 ** Only the descending mass experiences an increase in translational KE. **
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RESPONSE --> ok
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08:40:07 Does any of the bolts attached to the Styrofoam wheel gain more kinetic energy than some other bolt? Explain.
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RESPONSE --> The bolts toward the outside of the wheel.
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08:40:11 ** The bolts toward the outside of the wheel are moving at a greater velocity relative to some fixed point, so their kinetic energy is greater since k = 1/2 m v^2 **
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RESPONSE --> ok
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08:41:10 What is the moment of inertia of the Styrofoam wheel and its bolts?
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RESPONSE --> The moment of inertia for the center of its mass is equal to its radias times angular velocity. Moment of inertia of a bolt is m r^2.
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08:41:14 ** The moment of inertia for the center of its mass=its radias times angular velocity. Moment of inertia of a bolt is m r^2, where m is the mass and r is the distance from the center of mass. The moment of inertia of the styrofoam wheel is .5 M R^2, where M is its mass and R its radius. The wheel with its bolts has a moment of inertia which is equal to the sum of all these components. **
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RESPONSE --> ok
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08:44:55 How do we determine the angular kinetic energy of of wheel by measuring the motion of the falling mass?
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RESPONSE --> The velocity of the part of the wheel around which it is wound divided by its radius will give you the angular velocity of the wheel.
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08:44:59 ** STUDENT ANSWER AND INSTRUCTOR CRITIQUE: The mass falls at a constant acceleration, so the wheel also turns this fast. INSTRUCTOR CRITIQUE: We don't use the acceleration to find the angular KE, we use the velocity. The acceleration, if known, can be used to find the velocity. However in this case what we are really interested in is the final velocity of the falling mass, which is equal to the velocity of the part of the wheel around which it is wound. If we divide the velocity of this part of the wheel by the its radius we get the angular velocity of the wheel. **
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RESPONSE --> ok
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